In how many ways can 21 books on Sanskrit and | vedclass

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(B) To ensure that no two French books are together,we use the gap method.First,arrange the $21$ Sanskrit books in a row. These $21$ books create $22$

Vedclass · Accepted Answer

$1540$

Vedclass · Answer

(B) To ensure that no two French books are together,we use the gap method.First,arrange the $21$ Sanskrit books in a row. These $21$ books create $22$ possible gaps (including the ends) where the French books can be placed.Number of ways to arrange $21$ Sanskrit books = $21!$.Number of ways to arrange $19$ French books in the $22$ available gaps = $P(22, 19)$.However,the question asks for the number of ways to place the books,implying the books of the same language are identical or the arrangement is based on positions. Given the options,we calculate the selection of gaps: ${}^{22}C_{19}$.${}^{22}C_{19} = {}^{22}C_{3} = \frac{22 	imes 21 	imes 20}{3 	imes 2 	imes 1} = 22 	imes 7 	imes 10 = 1540$.

In how many ways can $21$ books on Sanskrit and $19$ books on French be placed in a row on a shelf such that two books on French may not be together?

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